The magnetic induction due to an ideal solenoid is independent of

There are $50$ turns of a wire in every $1$ $cm$ length of a long solenoid. If $4$ $A$ current is flowing in the solenoid,the approximate value of the magnetic field along its axis at an internal point and at one end will be respectively:

$A$ current-carrying solenoid is placed vertically and a particle of mass $m$ with charge $Q$ is released from rest. The particle moves along the axis of the solenoid. If $g$ is the acceleration due to gravity, then the acceleration $(a)$ of the charged particle will satisfy:

If the magnetic field inside a solenoid is $B$,then the magnetic energy stored in it per unit volume is (where $c$ is the speed of light in vacuum and $\varepsilon_0$ is the permittivity of free space).

$A$ long solenoid carrying current $I_1$ produces a magnetic field $B_1$ along its axis. If the current is reduced to $20 \%$ and the number of turns per $cm$ is increased five times,then the new magnetic field $B_2$ is equal to:

$A$ toroid is a long coil of wire wound over a circular core. If $r$ and $R$ are the radii of the coil and toroid respectively,the coefficient of self-induction of the toroid is (The magnetic field in it is uniform and $R >> r$). ($N =$ number of turns of the coil and $\mu_{0} =$ permeability of free space)